Sans air resistance, in standard gravity (g=9.80665 m/s/s), an object will fall 19.6133m in two seconds. The slightly higher value in the puzzle thread can be accounted for by local discrepancies in the gravitational field - if the ground is denser here, the gravity will be higher, for example.

Now, it's hard to find data on the terminal velocity of a falling bear (Google fails me once again), and whether air resistance will have a significant effect... it is possible that a sufficiently high local gravity would be enough to overcome the air resistance (for instance, gravity is higher at the poles, around g=9.832 m/s/s according to WP, so Vesuvius's kneejerk "Polar bear" guess might work)... but it might not. Regardless, you would have better luck with a bald bear, as it would have less air resistance, and would thus be more likely to be able to make the fall in the prescribed time.

My guess is that we are supposed to conclude based on the exact value of g in the problem that there is gravity but no centrifugal force, and we must therefore be at a pole, making the bear a polar bear.

However, this strikes me as rather an unlikely scenario. Apart from the above-mentioned problems with air resistance, gravity also varies from place to place: see this list of comparative gravities in various cities. The variation found in that list, .04 m/s2, is more than 40 times greater than the centrifugal acceleration at the equator, which is less than .001 m/s2.So the idea that one could determine centrifugal force and thereby determine latitude by measuring the local strength of gravity is pretty farfetched. Edit: this is wrong. Gmalivuk correctly points out that I'm off by a factor of (2π)2, so centrifugal force is about the same magnitude as the differences between different cities on this list, and probably explains much of the differences.

While we're on the subject, does anyone know why standard gravity is defined to be exactly 9.80665? Is this precise value in any way physically meaningful? If it isn't, why don't we use the much simpler value of 9.8 for standard gravity?

Mmmm g would also be effected by being on top of a mountin (I dont know with out looking it up if thats a bigger diffence then being at the polls). I think the biggest facter would be air resitace and that would change based on the size of the bear so a better question would be how old is the bear. PS I like the bald argument

While we're on the subject, does anyone know why standard gravity is defined to be exactly 9.80665? Is this precise value in any way physically meaningful? If it isn't, why don't we use the much simpler value of 9.8 for standard gravity?

Spoiler:

Probably for the same reason that the speed of light isn't quite 3e8 m/s, and standard atmospheric pressure is over 1e5 Pa... and, for that matter, why an inch isn't 2.5cm... inertia, and the desire not to change a value too much when it gets redefined in terms of something else.

Presumably, when they decided to define standard gravity as a specific value in m/s/s, they just took the previous value for g, however that was defined, rounded it off to more digits than you'd ever really need, and used that.

Apart from the above-mentioned problems with air resistance, gravity also varies from place to place: see this list of comparative gravities in various cities. The variation found in that list, .04 m/s2, is more than 40 times greater than the centrifugal acceleration at the equator, which is less than .001 m/s2.

(edited for brevity)

Spoiler:

I believe that gives us the answer. The bear is in Paris (well... Northern France) since the only kind of bear present in Northern France is The Brown Bear that gives us the color.

Apart from the above-mentioned problems with air resistance, gravity also varies from place to place: see this list of comparative gravities in various cities. The variation found in that list, .04 m/s2, is more than 40 times greater than the centrifugal acceleration at the equator, which is less than .001 m/s2.

(edited for brevity)

Spoiler:

I believe that gives us the answer. The bear is in Paris (well... Northern France) since the only kind of bear present in Northern France is The Brown Bear that gives us the color.

Spoiler:

On what basis do you rule out black, the color of the black bear, whose habitat includes the vicinity of Vancouver?

phlip wrote:

Spoiler:

Probably for the same reason that the speed of light isn't quite 3e8 m/s, and standard atmospheric pressure is over 1e5 Pa... and, for that matter, why an inch isn't 2.5cm... inertia, and the desire not to change a value too much when it gets redefined in terms of something else.

Presumably, when they decided to define standard gravity as a specific value in m/s/s, they just took the previous value for g, however that was defined, rounded it off to more digits than you'd ever really need, and used that.

Apart from the above-mentioned problems with air resistance, gravity also varies from place to place: see this list of comparative gravities in various cities. The variation found in that list, .04 m/s2, is more than 40 times greater than the centrifugal acceleration at the equator, which is less than .001 m/s2.

(edited for brevity)

Spoiler:

I believe that gives us the answer. The bear is in Paris (well... Northern France) since the only kind of bear present in Northern France is The Brown Bear that gives us the color.

Spoiler:

On what basis do you rule out black, the color of the black bear, whose habitat includes the vicinity of Vancouver?

Bears in nature are color coded. A black one will apparently ruin your solution by just existing. Back to the drawing board then.

as well as deducing geographical information from the falling time (i.e. centrifugal forces, gravitational anomaly), the official solution also used other facts about the habitats of bears, the difficulties involved in digging a trap, and the motivations for trapping them...

But, I think the official solution makes a few incorrect assumptions, so I want to see what you guys come up with.

i'm not sure if it was said (quickly scanned spoilers and the prominant one was not the right answer) so i'll say what i think:

Spoiler:

Gravity accelerates an object downward at 9.80665m/s. This means in 2 seconds the bear SHOULD travel 19.6133

but the bear travels 19.617, a bit too fast for normal acceleration.

now, as most of you know, the world is not a perfect sphere. Instead, the earth is an oval, like a squished sphere from the top/bottom. This means that at the north/south pole, the acceleration is slightly more (exact number unsure). This increase in the pull of gravity would fit the 19.617m in 2 seconds, and that it'd make sense with the riddle for the ONLY bears in at the pole are white.

But what panda would fall out of a tree into a trap? Surely to get it to fall out of the tree, you'd need to at least tranquilize it first, and if you do that you don't need a trap. If the hypothetical trapper was just hoping that a panda would climb up into the tree above the trap and accidentally fall in, somebody tell him he's dreaming.

Use the quadratic formula:[math]\begin{eqnarray*}h&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\2h&=&-2{r_e}\pm\sqrt{4{r_e}^2-4{r_e}^2(1-\frac{g_{0}}{g_h})}\\2h&=&-2{r_e}\pm\sqrt{4{r_e}^2(\frac{g_{0}}{g_h})}\\2h&=&-2{r_e}\pm2{r_e}\sqrt{\frac{g_{0}}{g_h}}\\h&=&{r_e}(-1\pm\frac{\sqrt{g_0}}{\sqrt{g_h}})\\h&=&6371000\:\textrm{m}(-1\pm\frac{\sqrt{9.80665}}{\sqrt{9.8085}})\\h&=&6371000\:\textrm{m}(-1+0.99990568959365644)\\h&=&-600.85159881479683\:\textrm{m}\end{eqnarray*}[/math]

Use the quadratic formula:[math]\begin{eqnarray*}h&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\2h&=&-2{r_e}\pm\sqrt{4{r_e}^2-4{r_e}^2(1-\frac{g_{0}}{g_h})}\\2h&=&-2{r_e}\pm\sqrt{4{r_e}^2(\frac{g_{0}}{g_h})}\\2h&=&-2{r_e}\pm2{r_e}\sqrt{\frac{g_{0}}{g_h}}\\h&=&{r_e}(-1\pm\frac{\sqrt{g_0}}{\sqrt{g_h}})\\h&=&6371000\:\textrm{m}(-1\pm\frac{\sqrt{9.80665}}{\sqrt{9.8085}})\\h&=&6371000\:\textrm{m}(-1+0.99990568959365644)\\h&=&-600.85159881479683\:\textrm{m}\end{eqnarray*}[/math]

So the hole is 600 meters below sea level... oops.

OOPS: Are there bears in the dead sea?

Sea of Galilee is also 600 meters i think? A seabear? but wouldn't the water slow the fall...assuming it "falls" instead of "floats"

Use the quadratic formula:[math]\begin{eqnarray*}h&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\2h&=&-2{r_e}\pm\sqrt{4{r_e}^2-4{r_e}^2(1-\frac{g_{0}}{g_h})}\\2h&=&-2{r_e}\pm\sqrt{4{r_e}^2(\frac{g_{0}}{g_h})}\\2h&=&-2{r_e}\pm2{r_e}\sqrt{\frac{g_{0}}{g_h}}\\h&=&{r_e}(-1\pm\frac{\sqrt{g_0}}{\sqrt{g_h}})\\h&=&6371000\:\textrm{m}(-1\pm\frac{\sqrt{9.80665}}{\sqrt{9.8085}})\\h&=&6371000\:\textrm{m}(-1+0.99990568959365644)\\h&=&-600.85159881479683\:\textrm{m}\end{eqnarray*}[/math]

So the hole is 600 meters below sea level... oops.

OOPS: Are there bears in the dead sea?

Sea of Galilee is ALSO YOU GUYS: 600 meters i think? A seabear? but wouldn't the water slow the fall...assuming it "falls" instead of "floats"

Hey, this isn't on topic, but...

Since when does "a[imath][/imath]lso" get changed to "also" in posts?

It's actually kind of funny, since it's recursive (look at the above quote that I did not edit manually)...

Also, since when does "e[imath][/imath]dit:" get changed to "edit:"?

EDIT: And "skep[imath][/imath]tical" get changed to "skeptical". Am I really behind the times or did this all happen recently?

EDIT 2: This is the last time...

"intern[imath][/imath]et" -> "internet"

I'm amazed that I noticed all these in such a short amount of time. Is there a list somewhere, or a thread that keeps track of these?

I realise this thread is quite old, but I ran the numbers before I realised so I'll post this anyways.

I decided to test these conditions, so I derived a formula based on kinetic and gravitational potential energies. time_to_crash = sqrt(g*h/.5)/gNow that I've derived that, I can reorder to find g, and thus help narrow the location. g=2*h/t^2

This gives us 9.8085 m/s^2. We can now use this to find out altitude! We now take the law of universal gravitation, F=Gm1m2/r^2, reorder for altitude, r=sqrt(Gm1m2/F), run in the numbers, and we get 6374.0199574126711249253201224927 Kilometres.